Answer :
To address the given problem, we want to perform a Chi-square goodness-of-fit test to determine if the number of M&Ms is uniformly distributed across all six colors. Here's the detailed step-by-step solution:
1. Introduction:
- Sally is interested in verifying if the counts of M&Ms in a 1-pound bag are uniformly distributed across six different colors.
2. Data Information:
- The counts of M&Ms for each color are given:
- Brown: 70
- Red: 85
- Yellow: 81
- Blue: 79
- Orange: 93
- Green: 96
3. Null Hypothesis (H₀):
- The null hypothesis for a Chi-square goodness-of-fit test in this context evaluates whether the observed counts follow a uniform distribution.
- Therefore, the appropriate null hypothesis is:
[tex]\[ H_0: \text{The proportion of each color is } \frac{1}{6} \][/tex]
4. Explanation of the Hypothesis:
- Under the null hypothesis, it is assumed that the M&Ms are evenly distributed among the six colors. This implies that each color should constitute [tex]\(\frac{1}{6}\)[/tex] of the total M&Ms in the bag.
5. Calculation of Expected Counts:
- To assess this hypothesis, we need to compute the expected counts for each color based on the assumption of uniformity (proportional distribution of [tex]\(\frac{1}{6}\)[/tex]).
- The total number of M&Ms in all colors combined:
[tex]\[ \text{Total number of M&Ms} = 70 + 85 + 81 + 79 + 93 + 96 = 504 \][/tex]
- The expected count for each color can be calculated as:
[tex]\[ \text{Expected count for each color} = \text{Total number of M&Ms} \times \text{Expected proportion per color} = 504 \times \frac{1}{6} = 84 \][/tex]
- Hence, the expected counts are:
[tex]\[ [84, 84, 84, 84, 84, 84] \][/tex]
6. Results from the Observed Data:
- The observed counts from the table are:
[tex]\[ [70, 85, 81, 79, 93, 96] \][/tex]
- And we calculated that the expected counts under the null hypothesis are:
[tex]\[ [84, 84, 84, 84, 84, 84] \][/tex]
7. Conclusion:
- The null hypothesis we are testing with the Chi-square goodness-of-fit test is:
[tex]\[ H_0: \text{The proportion of each color is } \frac{1}{6} \][/tex]
- This hypothesis states that the M&Ms are uniformly distributed across the six colors, each having a proportion of [tex]\(\frac{1}{6}\)[/tex].
By stating this null hypothesis, you can proceed to carry out the Chi-square goodness-of-fit test to determine whether there is sufficient evidence to reject this hypothesis in favor of the alternative—that the proportions are not uniform across the colors.
1. Introduction:
- Sally is interested in verifying if the counts of M&Ms in a 1-pound bag are uniformly distributed across six different colors.
2. Data Information:
- The counts of M&Ms for each color are given:
- Brown: 70
- Red: 85
- Yellow: 81
- Blue: 79
- Orange: 93
- Green: 96
3. Null Hypothesis (H₀):
- The null hypothesis for a Chi-square goodness-of-fit test in this context evaluates whether the observed counts follow a uniform distribution.
- Therefore, the appropriate null hypothesis is:
[tex]\[ H_0: \text{The proportion of each color is } \frac{1}{6} \][/tex]
4. Explanation of the Hypothesis:
- Under the null hypothesis, it is assumed that the M&Ms are evenly distributed among the six colors. This implies that each color should constitute [tex]\(\frac{1}{6}\)[/tex] of the total M&Ms in the bag.
5. Calculation of Expected Counts:
- To assess this hypothesis, we need to compute the expected counts for each color based on the assumption of uniformity (proportional distribution of [tex]\(\frac{1}{6}\)[/tex]).
- The total number of M&Ms in all colors combined:
[tex]\[ \text{Total number of M&Ms} = 70 + 85 + 81 + 79 + 93 + 96 = 504 \][/tex]
- The expected count for each color can be calculated as:
[tex]\[ \text{Expected count for each color} = \text{Total number of M&Ms} \times \text{Expected proportion per color} = 504 \times \frac{1}{6} = 84 \][/tex]
- Hence, the expected counts are:
[tex]\[ [84, 84, 84, 84, 84, 84] \][/tex]
6. Results from the Observed Data:
- The observed counts from the table are:
[tex]\[ [70, 85, 81, 79, 93, 96] \][/tex]
- And we calculated that the expected counts under the null hypothesis are:
[tex]\[ [84, 84, 84, 84, 84, 84] \][/tex]
7. Conclusion:
- The null hypothesis we are testing with the Chi-square goodness-of-fit test is:
[tex]\[ H_0: \text{The proportion of each color is } \frac{1}{6} \][/tex]
- This hypothesis states that the M&Ms are uniformly distributed across the six colors, each having a proportion of [tex]\(\frac{1}{6}\)[/tex].
By stating this null hypothesis, you can proceed to carry out the Chi-square goodness-of-fit test to determine whether there is sufficient evidence to reject this hypothesis in favor of the alternative—that the proportions are not uniform across the colors.